The sum of the squares of two consecutive odd integers is 130. find the integers

N²+(n+2)²=130
Expand the binomial: n²+n²+4n+4 = 130
2n²+4n+4-130 = 0
Collect terms and put the equation in standard form: 2n²+4n-126=0
Factor out the 2 : n²+2n-63 =0
Now, you need to find two integers that have the following properties:

Their product is -63, and their sum is +2. Since the product is negative, we know that the two numbers are of opposite signs, and since their sum is positive, we know the larger of the two must be positive.

So, what are the factors of 63? Let's try 7 and 9. If we choose -7 and +9 we have a fit for the properties. -7 * 9= -63 and -7 + 9 = 2.
Therefore, the left hand expression factors to n - 7 and n + 9, so n = 7 or n=-9

That means that our first consecutive odd integer is either 7 or -9, so the second consecutive odd integer is either 9 or -7.